Find the area bounded by the curve nbsp y equals 4 minus x squared and the x axis between x equals 1 and x equals 6.
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Part 1
The area bounded by the curve nbsp y equals 4 minus x squared and the x axis between x equals 1 and x equals 6 is 
  

.
​(Type an integer or a simplified​ fraction.)

To find the area bounded by the curve \( y = 4 - x^2 \) and the x-axis between \( x = 1 \) and \( x = 6 \), we need to calculate the definite integral of \( y = 4 - x^2 \) with respect to \( x \) over the interval \([1, 6]\).

The formula for the area under the curve is:

\[
A = \int_{1}^{6} (4 - x^2) \, dx
\]

### Step 1: Integrate the function

First, we'll compute the integral of \( 4 - x^2 \):

\[
\int (4 - x^2) \, dx = 4x - \frac{x^3}{3} + C
\]

### Step 2: Apply the limits of integration

Now, we evaluate the integral from 1 to 6:

\[
A = \left[ 4x - \frac{x^3}{3} \right]_{1}^{6}
\]

Substituting the upper limit (\(x = 6\)):

\[
A_6 = 4(6) - \frac{6^3}{3} = 24 - \frac{216}{3} = 24 - 72 = -48
\]

Substituting the lower limit (\(x = 1\)):

\[
A_1 = 4(1) - \frac{1^3}{3} = 4 - \frac{1}{3} = \frac{12}{3} - \frac{1}{3} = \frac{11}{3}
\]

### Step 3: Calculate the total area

Now, subtract the value at \(x = 1\) from the value at \(x = 6\):

\[
A = -48 - \frac{11}{3} = \frac{-144}{3} - \frac{11}{3} = \frac{-155}{3}
\]

Since the area cannot be negative, we take the absolute value:

\[
A = \frac{155}{3} \, \text{square units}
\]

Thus, the area bounded by the curve and the x-axis is \( \frac{155}{3} \) square units.

Would you like further details on any step or have any questions?

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Here are 5 related questions:
1. How do you interpret the meaning of a negative result in definite integrals?
2. What is the geometric significance of an area calculation under a curve?
3. How would you calculate the area if the curve were above and below the x-axis?
4. What is the difference between definite and indefinite integrals in terms of area?
5. How do you calculate the area between two curves?

**Tip:** Always check whether the function crosses the x-axis within the given limits, as it may affect the area calculation.

Find the area bounded by the curve y = 4 - x^2 and the x-axis between x = 1 and x = 6. (Type an integer or a simplified fraction.)

Learn how to find the area bounded by the curve y = 4 - x^2 and the x-axis between x = 1 and x = 6 using definite integration. This problem involves calculus concepts, such as the definite integral and the application of the fundamental theorem of calculus. Suitable for students in Grades 11-12 or early college.

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